Problem: Consider the polar curve $r=2\cos(\theta)$ for $0\le \theta < \pi$. At which values of $\theta$ does the graph of $r$ have a horizontal tangent line? Choose 1 answer: Choose 1 answer: (Choice A) A $0$ only (Choice B) B $0$ or $\dfrac{\pi}{2}$ (Choice C) C $\dfrac{\pi}{4}$ only (Choice D) D $\dfrac{\pi}{4}$ or $\dfrac{3\pi}{4}$
Explanation: A horizontal line has a slope of $0$. The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Once we have an expression for the slope of the tangent line, we can look for the $\theta$ -values that make $\dfrac{dy}{d\theta}=0$ but don't make $\dfrac{dx}{d\theta}=0$ (because then $\dfrac{dy}{dx}$ will be undefined). For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={2\cos(\theta)}\cos(\theta) \\\\ &=2\cos^2(\theta) \\\\ y&={2\cos(\theta)} \sin(\theta) \\\\ &=\sin(2\theta) \end{aligned}$ Let's find $\dfrac{dy}{d\theta}$ and $\dfrac{dx}{d\theta}$. $\begin{aligned} y(\theta)&=\sin(2\theta) \\\\ \dfrac{dy}{d\theta}&=2\cos(2\theta) \\\\ \\\\ x(\theta)&=2\cos^2(\theta) \\\\ \dfrac{dx}{d\theta}&=-2\cdot2\cos(\theta)\sin(\theta) \\\\ &=-2\sin(2\theta) \end{aligned}$ Now let's solve $\dfrac{dy}{d\theta}=0$ on the interval $0\le \theta < \pi$. $\begin{aligned} \dfrac{dy}{d\theta}&=0 \\\\ 2\cos(2\theta)&=0 \\\\ \cos(2\theta)&=0 \\\\ 2\theta=\dfrac{\pi}{2}+k\cdot\pi&\text{, where }k\text{ is an integer} \\\\ \theta=\dfrac{\pi}{4}+k\dfrac{\pi}{2}&\text{, where }k\text{ is an integer} \end{aligned}$ Within our interval, our possibilities are $\theta=\dfrac{\pi}{4}$ and $\theta=\dfrac{3\pi}{4}$. Finally, we evaluate $\dfrac{dx}{d\theta}$ for our two possible values of $\theta$ and require that $\dfrac{dx}{d\theta}\ne0$. $\begin{aligned} \left.\dfrac{dx}{d\theta} \right| _{{\theta =\tfrac{\pi }{4}}}&=-2\sin\left(2\cdot{\dfrac{\pi}{4}}\right) \\\\ &=-2\sin\left(\dfrac{\pi}{2}\right) \\\\ &=-2 \\\\ \\\\ \left.\dfrac{dx}{d\theta} \right| _{{\theta =\tfrac{3\pi }{4}}}&=-2\sin\left(2\cdot{\dfrac{3\pi}{4}}\right) \\\\ &=-2\sin\left(\dfrac{3\pi}{2}\right) \\\\ &=2 \end{aligned}$ The graph of $r$ has a horizontal tangent at $\theta=\dfrac{\pi}{4}$ or $\theta=\dfrac{3\pi}{4}$. The graphs of the tangents are shown. ${1}$ ${2}$ ${0}$ ${\frac{1}{8}\pi}$ ${\frac{1}{4}\pi}$ ${\frac{3}{8}\pi}$ ${\frac{1}{2}\pi}$ ${\frac{5}{8}\pi}$ ${\frac{3}{4}\pi}$ ${\frac{7}{8}\pi}$ ${\pi}$ ${\frac{9}{8}\pi}$ ${\frac{5}{4}\pi}$ ${\frac{11}{8}\pi}$ ${\frac{3}{2}\pi}$ ${\frac{13}{8}\pi}$ ${\frac{7}{4}\pi}$ ${\frac{15}{8}\pi}$